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Talk:0
While the infinities can be larger and larger, can zeroes be smaller and smaller too? When we write 0, we usually mean that it is 0.000... (with \(\aleph_0\) zeroes), i.e. it is \(1 \over \aleph_0\). But can exist even smaller zero which is represented by \(1 \over \aleph_1\), in other words, having uncountably many zeroes after decimal point? Ikosarakt1 (talk ^ ) 10:52, September 9, 2013 (UTC) I hope it is obvious that this is not the case for real numbers. In any group, If we had two numbers which we claim to be zeroes, by multiplying them we would prove they are equal. However, if we talk about non-zero infinitesimals, we have inverses of any ordinals, e.g. in surreal numbers. LittlePeng9 (talk) 12:37, September 9, 2013 (UTC) :Is it true that zero actually has absolute infinity of zeroes, not just \(\aleph_0\)? Ikosarakt1 (talk ^ ) 15:57, September 9, 2013 (UTC) :Real numbers have only countably many decimal digits, so I believe answer is no. LittlePeng9 (talk) 16:38, September 9, 2013 (UTC) :There have been people who thought they overthrew Cantor's diagonalization argument by misunderstanding the whole \(\aleph_0\)-digits-after-the-decimal-point ordeal. Infinity is so counterintuitive that it can even get political. FB100Z • talk • 03:02, September 10, 2013 (UTC) How about an idea that irrational numbers like pi actually have digits on transfinite positions? Say, pi can be in form 3.141592...56319495... (where the second "5" has \(\omega\)-th position). Ikosarakt1 (talk ^ ) 13:36, September 10, 2013 (UTC) :This idea would make perfect sense in continuum containing infinitesimals, i.e. reciprocals of ordinals. Pi is real, not having transfinite places, but idea would work. For many (let me call them like that here) superreals suffiscently far in expansion they'd have absolute infinity of zeroes. But we can give concept of superirrational, not having this property. LittlePeng9 (talk) 13:57, September 10, 2013 (UTC) :The usual real numbers don't work that way, but there's nothing to stop us from defining a new system that allows it. You could do things like \(1/\omega = 1/10^\omega = 0.000...1\). Note that the "number" of digits after the decimal point has to be a countable ordinal; otherwise we can't write out all the digits in sequence. :As an added note, I challenge people who think that 0.999... ≠ 1 by asking them what 1 - 0.999... is. They usually answer "the smallest real number" or 0.000...1. I will clarify that 1 - 0.999... = 0 even if we introduce the hyperreals — the reals are closed over addition and both 1 and -0.999... are valid reals. FB100Z • talk • 17:56, September 10, 2013 (UTC) :Note that, if we worked in surreals, 0.999... could be different from 1 (it isn't, but this is hypothetical example of why), because all 0, 0.9, 0.99, 0.999,... are smaller than \(1-\varepsilon\), one minus infinitesimal. That's because least upper bound property fails for surreals, and limits are, actually, undefined in standard sense. LittlePeng9 (talk) 19:47, September 10, 2013 (UTC) ::I guess it's all a matter of convention. If the surreals are an extension of the real numbers, then 1 = 0.999... as it normally is. If the surreals are a modification and extension of the real numbers, then we may or may not have 1 ≠ 0.999... It's really confusing now that I've thought it through a bit more :/ FB100Z • talk • 19:52, September 10, 2013 (UTC) ::The question what zero is: just the lim(1,0.1,0.01,0.001,...) or something yet smaller interests me in last days. If we consider zero as an real number, then it must be the first variant. Otherwise, it could be even the inverse of the Absolute Infinity. I thought that hyperreals mess up limits; now I realize that they're just another way of defining them. For example, define the st function as \(\text{st}(x + \epsilon) = \text{st}(x - \epsilon) = \text{st}(x) = x\) for real x and infinitesimal \(\epsilon\). Then we can define \(f'(x) = \text{st}\left(\frac{f(x + \epsilon) - f(x)}{\epsilon}\right)\), equivalent to the traditional \(f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}\). FB100Z • talk • 02:39, September 12, 2013 (UTC) Absolute values The thing is that "| 0 | is the only absolute value expression that equals itself" is incorrect. This holds for other positive real numbers as well, for example |1| equals 1, so it equals itself. |1.5| equals 1.5, so it equals itself. Since this property isn't just limited to 0, I don't think we actually need to include it on the article. -- ☁ I want more ⛅ 04:10, February 18, 2017 (UTC) Very true! I ment 0 is the only number where itself and its opposite equal the same thing: 0. ( |±0| = |±0|) Nathan da' R. 17:28, February 18, 2017 (UTC) Whole numbers The article says "0 is the only whole number that is not a natural number. Whole numbers include 0, 1, 2, 3, 4, etc. while naturals numbers skip zero and continue: 1, 2, 3, 4, 5, etc." However some websites like this one says that some people use the term "whole numbers" to mean integers. I think this paragraph should be deleted. Rpakr (talk) 20:14, January 11, 2018 (UTC) : I second that Nathan Richardson "Simon Weston" 00:51, January 12, 2018 (UTC)